The critical Points of f(x,y)=x3−3x+3y2−6y are:

Question

crocolylec · Accepted Answer

Step 1
The critical points of a function f(x, y) can be obtained by solving the partial derivatives

\frac{\partial f}{\partial x} = 0 and \frac{\partial f}{\partial y} = 0

.
The given function is

f (x, y) = x^{3} - 3 x + 3 y^{2} - 6 y

.
Evaluate

\frac{\partial f}{\partial x} = 0

as follows.

\frac{\partial f}{\partial x} = 0

\frac{\partial}{\partial x} (x^{3} - 3 x + 3 y^{2} - 6 y) = 0

3 x^{2} - 3 + 0 - 0 = 0

3 x^{2} = 3

x^{2} = 1

x=-1 and x=1
Thus, the values of x are -1 and 1.
Step 2
Evaluate

\frac{\partial f}{\partial y} = 0

as follows.

\frac{\partial f}{\partial y} = 0

\frac{\partial}{\partial y} (x^{3} - 3 x + 3 y^{2} - 6 y) = 0

0+0+6y-6=0
6y=6
y=1
Thus, the value of y is 1.
Therefore, the critical points of the given function are (-1, 1) and (1,1).

The critical Points of f(x,y)=x^{3}-3x+3y^{2}-6y are